If `n lt p lt 2n` and p is prime and `N=.^(2n)C_(n)`, then

Options

A. n divides N
B. p does not divide N
C. `p^(2)` divides N
D. `p^(2)` does not divide N

Answer

To solve the problem, we need to analyze the given information step by step.

### Step 1: Understand the Given Information
We are given that \( n < p < 2n \) and \( p \) is a prime number. We also have \( N = \binom{2n}{n} \).

### Step 2: Write the Formula for \( N \)
The formula for \( N \) can be expressed as:
\[ N = \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \]

### Step 3: Factorial Expansion
We can expand \( (2n)! \) as follows:
\[ (2n)! = (2n)(2n-1)(2n-2)\ldots(n+1)(n!) \]
Thus, we can rewrite \( N \):
\[ N = \frac{(2n)(2n-1)(2n-2)\ldots(n+1)(n!)}{(n!)^2} = \frac{(2n)(2n-1)(2n-2)\ldots(n+1)}{n!} \]

### Step 4: Analyze the Prime \( p \)
Since \( p \) is a prime number and falls between \( n \) and \( 2n \), we need to check how many times \( p \) appears in the numerator and denominator of \( N \).

- In the numerator, \( p \) can appear in the product \( (2n)(2n-1)\ldots(n+1) \).
- In the denominator, \( p \) appears in \( n! \).

### Step 5: Count the Occurrences of \( p \)
Since \( p \) is greater than \( n \) but less than \( 2n \), it will appear exactly once in the numerator (as \( p \) is included in the range from \( n+1 \) to \( 2n \)) and not at all in the denominator (as \( n! \) only includes numbers up to \( n \)).

### Step 6: Conclusion on Divisibility
Since \( p \) appears exactly once in the numerator and not at all in the denominator, it implies that \( p \) divides \( N \) completely.

However, since \( p \) is a prime number and it does not appear in the denominator, we conclude that:
\[ p \text{ divides } n \text{ completely.} \]

### Final Answer
The correct relation is that \( p \) divides \( n \).

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